Question
If $\text{f(x)}=\text{x}^3-\frac{1}{\text{x}^3},$ show that $\text{f(x)}+\text{f}\Big(\frac{1}{\text{x}}\Big)=0$

Answer

We have,
$\text{f(x)}=\text{x}^3-\frac{1}{\text{x}^3}\ ....(\text{i})$
Now, $\text{f}\Big(\frac{1}{\text{x}}\Big)=\Big(\frac{1}{\text{x}}\Big)^3-\frac{1}{\big(\frac{1}{\text{x}}\big)^3}$
$=\frac{1}{\text{x}^3}-\frac{1}{\frac{1}{\text{x}^3}}$
$\text{f}\Big(\frac{1}{\text{x}}\Big)=\frac{1}{\text{x}^3}-\text{x}^3\ ....(\text{ii})$
Adding equation (i) and equation (ii), we get
$\text{f(x)}+\text{f}\Big(\frac{1}{\text{x}}\Big)=\Big(\text{x}^3-\frac{1}{\text{x}^3}\Big)+\Big(\frac{1}{\text{x}^3}-\text{x}^3\Big)$
$=\text{x}^3-\frac{1}{\text{x}^3}+\frac{1}{\text{x}^3}-\text{x}^3$
$=0$
$\therefore\ \text{f(x)}+\text{f}\Big(\frac{1}{\text{x}}\Big)=0$ Hence, proved.

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